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3 years ago

Scintillatium has a halflife of 16 minutes. if a sample has 800 grams, find a formula for its mass after t minutes.

1 answer:

8 0

Halflife is the time needed for a radioactive molecule to decay half of its current mass. If t is the time elapsed, the formula for the halflife would be:
final mass= original mass * $\frac{1}{2} ^{\frac{t}{halflife}$

If you put the information of the problem into the formula, the equation will be:
final mass= original mass * $\frac{1}{2} ^{\frac{t}{halflife}$
final mass= 800g * $\frac{1}{2} ^{\frac{t}{16 min}$

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Answer:

1. A. The plant leans toward window.

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7. The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430°C. At the start of the reaction there are

Juli2301 [7.4K]

Answer:

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] = 0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

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Number of moles HI = 0.886 moles

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Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Q =(n(HI)²) /(nH2 *nI2)

Q = 0.886²/(0.714*0.984)

Q =1.117

Q<Kc This means the reaction goes to the right (side of products)

Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 -X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Kc =(n(HI)²) /(nH2 *nI2)

KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

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